HW due tonight Please Help! I think I figured some out.

boydoineedhelp

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Oct 25, 2009
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I tried both, but I'm not sure I'm doing either right. I don't know what to do for 2b) at all. I have to show the breakdowns on all my work. My homework is due tonight. Please help!

1. Using Binomial Distribution Formula P(x)= [sub:1sipkakk]n[/sub:1sipkakk] C [sub:1sipkakk]x[/sub:1sipkakk] P [sup:1sipkakk]x[/sup:1sipkakk] q [sup:1sipkakk](n - x)[/sup:1sipkakk]

An unfair coin (weighted) is flipped 3 times. The probability of getting a head is . Use the Binomial Formula to calculate the following (keep your answers as fractions or round to 3 decimal places):

(a) The probability of getting exactly three heads. SW

P(3)= [sub:1sipkakk]3[/sub:1sipkakk] C [sub:1sipkakk]3[/sub:1sipkakk] x .42 [sup:1sipkakk]3[/sup:1sipkakk] x (1-.42) [sup:1sipkakk]3-3[/sup:1sipkakk]

= 1 x .42 [sup:1sipkakk]3[/sup:1sipkakk] x .58 [sup:1sipkakk]0[/sup:1sipkakk]
= .074088
= .0741


USE THE PROBABILITY IN (A) ABOVE TO CALCULATE THE FOLLOWING:

(b) The probability of getting fewer than three heads. SW

P = 1 - .0741
= .9259


2. Binomial Distribution Table

Assume that 85% of the cars on a freeway are traveling faster than 75 miles per hour. A random sample of 15 cars was observed under normal driving conditions with no police car in sight.

(a) What is the probability that exactly 14 of them were going faster than 75 miles per hour? SW

n = 15, p = .85, x = 14

P(14) = [sub:1sipkakk]15[/sub:1sipkakk] C [sub:1sipkakk]14[/sub:1sipkakk] x .85 [sup:1sipkakk]14[/sup:1sipkakk] x (1 - .85) [sup:1sipkakk]15-14[/sup:1sipkakk]

= 15 x .85 [sup:1sipkakk]14[/sup:1sipkakk] x .15 [sup:1sipkakk]1[/sup:1sipkakk]
= .2312
=. 231

(b) What is the probability that 10 or more of them were going faster than 75 miles per hour? SW
 

soroban

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Jan 28, 2005
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5,588
Hello, boydoineedhelp!

I'd say you're doing fine!


\(\displaystyle 1. \text{Binomial Distribution Formula: }\;P(x) \;=\; _nC_x\,p^xq^{n-x}\)

An unfair coin (weighted) is flipped 3 times. The probability of getting a head is 0.85.
Use the Binomial Formula to calculate the following. (Round to 3 decimal places.)

(a) The probability of getting exactly three heads. SW

\(\displaystyle P(3H) \:=\: _3C_3(0.42)^3(0.58)^0 \:=\:0.074088 \:\approx\:0.074\)


(b) The probability of getting fewer than three heads. SW

\(\displaystyle P(H < 3) \:=\: 1 - 0.074 \:=\:0.926\)

All correct!




2. Assume that 85% of the cars on a freeway are traveling faster than 75 mph.
A random sample of 15 cars was observed under normal driving conditions with no police car in sight.

(a) What is the probability that exactly 14 of them were going faster than 75 mph? SW

\(\displaystyle n = 15,\:p = 0.85,\: q = 0.15, \:x = 14\)

\(\displaystyle P(14) \:= \:_{15}C_{14} (0.85)^{14}(0.15)^1 \:=\: 0.231231756 \:\approx\:0.231\)

Right!



(b) What is the probability that 10 or more of them were going faster than 75 mph?

\(\displaystyle \text{We must find the probabilitites for }x \:=\:10, 11, 12, 13, 14, 15\:\text{ and add them up.}\)

. . \(\displaystyle \begin{array}{ccccc} P(10) ^&=& _{15}C_{10}(0.85)^{10}(0.15)^5 &=& 0.044895302 \\ P(11) &=& _{15}C_{11}(0.85)^{11}(0.15)^4 &=& 0.115630411 \\ P(12) &=& _{15}C_{12}(0.85)^{12}(0.15)^3 &=& 0.218429998 \\ P(13) &=& _{15}C_{13}(0.85)^{13}(0.15)^2 &=& 0.285639229 \\P(14) &=& _{15}C_{14}(0.85)^{14}(0.15)^1 &=& 0.231231756 \\ P(15) &=& _{15}C_{15}(0.85)^{15}(0.15)^0 &=& 0.087354219 \\ & & & & ------ \\ & & & & 0.984275718 \end{array}\)

\(\displaystyle \text{Therefore: }\;P(\geq 10) \;\approx\;0.984\)

 
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